Problem: $ D = \left[\begin{array}{rr}4 & -1 \\ 5 & -2 \\ 2 & 3\end{array}\right]$ $ F = \left[\begin{array}{rr}-2 & 4 \\ -2 & 2\end{array}\right]$ What is $ D F$ ?
Explanation: Because $ D$ has dimensions $(3\times2)$ and $ F$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ D F = \left[\begin{array}{rr}{4} & {-1} \\ {5} & {-2} \\ \color{gray}{2} & \color{gray}{3}\end{array}\right] \left[\begin{array}{rr}{-2} & \color{#DF0030}{4} \\ {-2} & \color{#DF0030}{2}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ D$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ D$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ D$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{4}\cdot{-2}+{-1}\cdot{-2} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ D$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{-2}+{-1}\cdot{-2} & ? \\ {5}\cdot{-2}+{-2}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ D$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{-2}+{-1}\cdot{-2} & {4}\cdot\color{#DF0030}{4}+{-1}\cdot\color{#DF0030}{2} \\ {5}\cdot{-2}+{-2}\cdot{-2} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{4}\cdot{-2}+{-1}\cdot{-2} & {4}\cdot\color{#DF0030}{4}+{-1}\cdot\color{#DF0030}{2} \\ {5}\cdot{-2}+{-2}\cdot{-2} & {5}\cdot\color{#DF0030}{4}+{-2}\cdot\color{#DF0030}{2} \\ \color{gray}{2}\cdot{-2}+\color{gray}{3}\cdot{-2} & \color{gray}{2}\cdot\color{#DF0030}{4}+\color{gray}{3}\cdot\color{#DF0030}{2}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}-6 & 14 \\ -6 & 16 \\ -10 & 14\end{array}\right] $